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VERY simple Velocity vs time graph interpretation

  1. Jan 25, 2008 #1
    Wow, I'm extremely frustrated with this one.

    1. The problem statement, all variables and given/known data

    http://img262.imageshack.us/img262/4616/probre1.jpg

    I forgot to label the axis's. The X is time and the Y is Velocity

    3. The attempt at a solution

    Iv done a few things on this one:

    First, I figured I would just count the area under it but I submitted the answer and it was wrong

    Second, I used x = x(o) + v(o)t plugging in the velocity and time from the graph along with the x(o) which is given in the problem

    Any help is appriciated, Thank You
     
  2. jcsd
  3. Jan 25, 2008 #2
    Last edited: Jan 25, 2008
  4. Jan 25, 2008 #3
    See, I did that and counted the area under the positive area in the first quadrant and then I subtracted it from the negative area in the 4th quad. I got 29 for the positive area and 6 for the negative, giving me 23 and it was wrong. So I figured maybe I had to add the areas for 35 but that as well came out wrong. :confused:
     
  5. Jan 25, 2008 #4
    23m is the displacement. You are asked for the position.
     
  6. Jan 25, 2008 #5
    ughh, 1 more hint, im not getting it
     
  7. Jan 25, 2008 #6
    If the initial position is [itex]x(0)=-10 m[/itex] and the displacement is 15 m, what's the final position?
     
  8. Jan 25, 2008 #7
    x(0) is 10. I'm confused how the position is different from the displacement. I know the displacement is how far away it is from the orgin no matter how it travels but the position would be. I'm just not seeing it .
     
  9. Jan 25, 2008 #8
    You are initial at [itex]x(0)=10 m[/itex]. If your displacement [itex]\Delta x[/itex] is -10 m you should arrive to the origin.

    [tex]\Delta x=x_f-x_i\Rightarrow x_f=\Delta x+x_i[/tex]

    Do you understand that?
     
  10. Jan 25, 2008 #9
    I'm sorry, This isn't making sense to me, I'm usually good with graph interpretation.
     
  11. Jan 25, 2008 #10
    Ok, let's try this.

    The initial position is 12 m. You walk 4 m at the positive direction. What is your displacement and what is your final position?
     
  12. Jan 25, 2008 #11
    displacement would be 4 m and your final position is 16 m. I see where your going with this so I would simply add my displacement to the initial position? make the answer 33?
     
  13. Jan 25, 2008 #12
  14. Jan 25, 2008 #13
    You know what it was? I was thinking of the X axis as if it were distance even though I KNEW it was time. Because of this I was thinking its position was at 9 and from the initial position it would be -1 m . DUHHH. Thanks so much for your help.
     
  15. Jan 25, 2008 #14
    Always it helps if you draw a picture o the actuall movement!
    Glad I helped! :smile:
     
  16. Jan 26, 2008 #15
    When you take the area under a curve you're finding the change in what ever quantity you're asked for. You need some sort of condition if you want to find the exact value. Without knowing that x(0) = 10m the best you could say about the graph would be the object moved 23m but I can't tell you EXACTLY where it is or EXACTLY where is started. Does that make sense?
     
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